Description
Si se tiene que
(x+2)3=x3+6x2+mx+n(x+2)^3 = x^3 + 6x^2 + mx + n (x+2)3=x3+6x2+mx+n
(x−5)3=x3−px2+qx−125(x-5)^3 = x^3 - px^2 + qx - 125 (x−5)3=x3−px2+qx−125
determine el valor de m+n+p+qm+n+p+qm+n+p+q.
A) 20 B) -15 C) 40 D) -10 E) 110
Si x+1x=23x + \frac{1}{x} = \sqrt[3]{2}x+x1=32, determine el valor de x6+1x3\frac{x^6+1}{x^3}x3x6+1.
A) 3−2333 - 2\sqrt[3]{3}3−233 B) 3−2233 - 2\sqrt[3]{2}3−232 C) 2−3232 - 3\sqrt[3]{2}2−332 D) 1−2331 - 2\sqrt[3]{3}1−233 E) 1−2231 - 2\sqrt[3]{2}1−232
Se tienen las dimensiones de un campo deportivo. Determine el área del gramado de fútbol si x=2003x = \sqrt[3]{200}x=3200. (Dimensiones: (x2+x+1)(x^2+x+1)(x2+x+1) m y (x−1)(x-1)(x−1) m)
A) 22003 m22\sqrt[3]{200} \text{ m}^223200 m2 B) 199 m2199 \text{ m}^2199 m2 C) 4003 m2\sqrt[3]{400} \text{ m}^23400 m2 D) 215 m2215 \text{ m}^2215 m2 E) 169 m2169 \text{ m}^2169 m2
Determine el equivalente reducido de M3\sqrt[3]{M}3M si
M=a6−b6(a+b)(a2−ab+b2)+b3;a≠−bM = \frac{a^6 - b^6}{(a+b)(a^2 - ab + b^2)} + b^3; \quad a \neq -b M=(a+b)(a2−ab+b2)a6−b6+b3;a=−b
A) aaa B) 0 C) -2 D) bbb E) 2
Si se cumple que
{(a+1)3+(b−1)3=18a+b=3\begin{cases} (a+1)^3 + (b-1)^3 = 18 \\ a+b=3 \end{cases} {(a+1)3+(b−1)3=18a+b=3
Determine el valor de (a+1)(b−1)(a+1)(b-1)(a+1)(b−1).
A) 6 B) -9 C) 1 D) 2 E) -5
Si:
(1x)2013+(1y)2013+(1z)2013=1x2013+y2013+z2013\left(\frac{1}{x}\right)^{2013} + \left(\frac{1}{y}\right)^{2013} + \left(\frac{1}{z}\right)^{2013} = \frac{1}{x^{2013} + y^{2013} + z^{2013}} (x1)2013+(y1)2013+(z1)2013=x2013+y2013+z20131
x≠−yx \neq -yx=−y, x≠−zx \neq -zx=−z, y≠−zy \neq -zy=−z
reducir:
x4026−y4026z4026+x4026−z4026y4026+z4026−y4026x4026\frac{x^{4026} - y^{4026}}{z^{4026}} + \frac{x^{4026} - z^{4026}}{y^{4026}} + \frac{z^{4026} - y^{4026}}{x^{4026}} z4026x4026−y4026+y4026x4026−z4026+x4026z4026−y4026
A) 1 B) 2 C) -1 D) 0 E) -2
Sea aaa un número que verifica a2−7a−1=0a^2 - 7a - 1 = 0a2−7a−1=0. Determine:
P=513(a3−364)a1132a4+13P = \frac{\sqrt[3]{51}(a^3 - 364)\sqrt[3]{a^{11}}}{2\sqrt[3]{a^4+ 1} } P=23a4+1351(a3−364)3a11
A) 1 B) 2/3 C) 1/2 D) 1/3 E) 3
Siendo {a;b;c;d}⊂R+\{a; b; c; d\} \subset \mathbb{R}^+{a;b;c;d}⊂R+; con (a+b+c+d)(bcd+acd+abd+abc)=16abcd(a+b+c+d)(bcd+acd+abd+abc) = 16abcd(a+b+c+d)(bcd+acd+abd+abc)=16abcd
Calcule:
a4+b4+c4+d44a+b+c+d\frac{\sqrt[4]{a^4 + b^4 + c^4 + d^4}}{a+b+c+d} a+b+c+d4a4+b4+c4+d4
A) 2\sqrt{2}2 B) 1 C) 4 D) 22\frac{\sqrt{2}}{2}22 E) 24\frac{\sqrt{2}}{4}42
Si: {x;y;z}⊂R−{0}∧x+y+z=xyz+xzy+yzx\{x; y; z\} \subset \mathbb{R} - \{0\} \land x+y+z = \frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x}{x;y;z}⊂R−{0}∧x+y+z=zxy+yxz+xyz
Halle el valor de:
((xy)7+(xz)7+(yz)7)(xyz)21x77+y77+z77\frac{((xy)^7 + (xz)^7 + (yz)^7)(xyz)^{21}}{x^{77} + y^{77} + z^{77}} x77+y77+z77((xy)7+(xz)7+(yz)7)(xyz)21
A) −12-\frac{1}{2}−21 B) -1 C) 1 D) 13\frac{1}{3}31 E) 12\frac{1}{2}21
A partir de a4+b2c2a=b4+a2c2b=c4+a2b2c=−abc\frac{a^4 + b^2c^2}{a} = \frac{b^4 + a^2c^2}{b} = \frac{c^4 + a^2b^2}{c} = -abcaa4+b2c2=bb4+a2c2=cc4+a2b2=−abc siendo {a;b;c}⊂R−{0}\{a; b; c\} \subset \mathbb{R} - \{0\}{a;b;c}⊂R−{0}
[(a6+b6)(b6+c6)(c6+a6)(a3+b3)(b3+c3)(c3+a3)]aa+b+c\sqrt[a+b+c]{\left[\frac{(a^6+b^6)(b^6+c^6)(c^6+a^6)}{(a^3+b^3)(b^3+c^3)(c^3+a^3)}\right]^a} a+b+c[(a3+b3)(b3+c3)(c3+a3)(a6+b6)(b6+c6)(c6+a6)]a
A) a9a^9a9 B) b12b^{12}b12 C) a15a^{15}a15 D) abc E) a3a^3a3
Siendo x,y,zx, y, zx,y,z tres números reales diferentes de cero que verifican:
(x−y)2+(y−z)2+(z−x)2=(x+y−2z)2+(y+z−2x)2+(z+x−2y)2(x-y)^2 + (y-z)^2 + (z-x)^2 = (x+y-2z)^2 + (y+z-2x)^2 + (z+x-2y)^2 (x−y)2+(y−z)2+(z−x)2=(x+y−2z)2+(y+z−2x)2+(z+x−2y)2
Simplifique:
x11+y11+z11+11x3y3z3(xy+yz+zx)xyz(x4+y4+z4)2\frac{x^{11} + y^{11} + z^{11} + 11x^3y^3z^3(xy+yz+zx)}{xyz(x^4+y^4+z^4)^2} xyz(x4+y4+z4)2x11+y11+z11+11x3y3z3(xy+yz+zx)
A) 1 B) 2 C) 3 D) 4 E) 5
Si x+y+z=1x+y+z=1x+y+z=1 halle el valor de
x3+y3+z3+3(xy+xz+yz)−1xyz\frac{x^3+y^3+z^3+3(xy+xz+yz)-1}{xyz} xyzx3+y3+z3+3(xy+xz+yz)−1
A) 1 B) 2 C) 3 D) 12\frac{1}{2}21 E) 13\frac{1}{3}31
Si 1x+y+1y+z+1z+x=0.5x+y+z\frac{1}{x+y} + \frac{1}{y+z} + \frac{1}{z+x} = \frac{0.5}{x+y+z}x+y1+y+z1+z+x1=x+y+z0.5
64(x+y+z)6−(x+y)6−(y+z)6−(z+x)6(x+y)3(y+z)3+(y+z)3(z+x)3+(z+x)3(x+y)3\frac{64(x+y+z)^6 - (x+y)^6 - (y+z)^6 - (z+x)^6}{(x+y)^3(y+z)^3 + (y+z)^3(z+x)^3 + (z+x)^3(x+y)^3} (x+y)3(y+z)3+(y+z)3(z+x)3+(z+x)3(x+y)364(x+y+z)6−(x+y)6−(y+z)6−(z+x)6
A) 2 B) -3 C) -2 D) 3 E) 6
Dadas las relaciones:
a=(a−b)2+b(a+1)a = (a-b)^2 + b(a+1) a=(a−b)2+b(a+1)
b=(b−c)2+c(b+1)b = (b-c)^2 + c(b+1) b=(b−c)2+c(b+1)
c=(c−a)2+a(c+1)c = (c-a)^2 + a(c+1) c=(c−a)2+a(c+1)
(a6−b6)2c6−4a3b3\frac{(a^6-b^6)^2}{c^6 - 4a^3b^3} c6−4a3b3(a6−b6)2
A) a3b3a^3b^3a3b3 B) b3c3b^3c^3b3c3 C) a3+b6a^3+b^6a3+b6 D) a3c3a^3c^3a3c3 E) c6c^6c6
Suponiendo que: (x22y)3+(y22x)3=3(x3+y3)\left(\frac{x^2}{2y}\right)^3 + \left(\frac{y^2}{2x}\right)^3 = 3(x^3+y^3)(2yx2)3+(2xy2)3=3(x3+y3) ; x6−y6=6x4y4⋅x3+y33x^6-y^6 = 6x^4y^4 \cdot \sqrt[3]{x^3+y^3}x6−y6=6x4y4⋅3x3+y3
x−3−y−33\frac{x^{-3}-y^{-3}}{3} 3x−3−y−3
A) 1 B) −23-\frac{2}{3}−32 C) -3 D) 23\frac{2}{3}32 E) −13-\frac{1}{3}−31
A partir de: a2+1a2+b2+1b2+c2+1c2=2(1a+1b+1c)\frac{a^2+1}{a^2} + \frac{b^2+1}{b^2} + \frac{c^2+1}{c^2} = 2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)a2a2+1+b2b2+1+c2c2+1=2(a1+b1+c1)
Donde {a;b;c}⊂R−{0}\{a; b; c\} \subset \mathbb{R} - \{0\}{a;b;c}⊂R−{0}. Calcule
(a3+b3+c3)3(a6+b6+c6)6(a9+b9+c9)9\frac{(a^3+b^3+c^3)^3(a^6+b^6+c^6)^6}{(a^9+b^9+c^9)^9} (a9+b9+c9)9(a3+b3+c3)3(a6+b6+c6)6
A) 1 B) 13\frac{1}{3}31 C) 3 D) 27 E) 127\frac{1}{27}271
Sabiendo que:
A=(x+y−1)3+3(x+y)(x+y−1)A = (x+y-1)^3 + 3(x+y)(x+y-1) A=(x+y−1)3+3(x+y)(x+y−1)
B=(x+y+1)3−3(x+y)(x+y+1)B = (x+y+1)^3 - 3(x+y)(x+y+1) B=(x+y+1)3−3(x+y)(x+y+1)
C=[(x+y)3+1]2+[(x+y)3−1]2C = \left[(x+y)^3+1\right]^2 + \left[(x+y)^3-1\right]^2 C=[(x+y)3+1]2+[(x+y)3−1]2
¿A qué es igual 2AB−C2AB-C2AB−C?
A) 4 B) 2 C) -4 D) -2 E) 0
Tres números reales diferentes a,ba, ba,b y ccc verifican la siguiente condición:
a=p+qa3;b=p+qb3;c=p+qc3a = \sqrt[3]{p+qa} \quad ; \quad b = \sqrt[3]{p+qb} \quad ; \quad c = \sqrt[3]{p+qc} a=3p+qa;b=3p+qb;c=3p+qc
Determine
a5+b5+c5pq\frac{a^5+b^5+c^5}{pq} pqa5+b5+c5
A) -5 B) 9 C) 23\frac{2}{3}32 D) 6 E) -6
Si x+y+z=1∧xy+yz+xz=xyzx+y+z=1 \land xy+yz+xz=xyzx+y+z=1∧xy+yz+xz=xyz. Calcule:
(x9+y9+z9)(1x11+y11+z11+1)(x^9+y^9+z^9)\left(\frac{1}{x^{11}+y^{11}+z^{11}+1}\right) (x9+y9+z9)(x11+y11+z11+11)
A) 1 B) 2 C) 12\frac{1}{2}21 D) 13\frac{1}{3}31 E) 3
Si a+b+c+d=0a+b+c+d=0a+b+c+d=0. Calcule
abc+abd+acd+bcda3+b3+c3+d3\frac{abc+abd+acd+bcd}{a^3+b^3+c^3+d^3} a3+b3+c3+d3abc+abd+acd+bcd
A) 1 B) 12\frac{1}{2}21 C) 2 D) 13\frac{1}{3}31 E) 19\frac{1}{9}91
Si se cumple:
ab+c+d+ba+c+d+ca+b+d+da+b+c=1\frac{a}{b+c+d} + \frac{b}{a+c+d} + \frac{c}{a+b+d} + \frac{d}{a+b+c} = 1 b+c+da+a+c+db+a+b+dc+a+b+cd=1
a2b+c+d+b2a+c+d+c2a+b+d+d2a+b+c\frac{a^2}{b+c+d} + \frac{b^2}{a+c+d} + \frac{c^2}{a+b+d} + \frac{d^2}{a+b+c} b+c+da2+a+c+db2+a+b+dc2+a+b+cd2
A) 0 B) 1 C) 2 D) 4 E) 14\frac{1}{4}41
Calcule el valor de ZZZ, si:
Zp+qp−q=12[a2−b2a2+b2][x1p+x1q]Z^{\frac{p+q}{p-q}} = \frac{1}{2}\left[\frac{a^2-b^2}{a^2+b^2}\right]\left[x^{\frac{1}{p}} + x^{\frac{1}{q}}\right] Zp−qp+q=21[a2+b2a2−b2][xp1+xq1]
cuando x=[a+ba−bp−q]2pqx = \left[\sqrt[p-q]{\frac{a+b}{a-b}}\right]^{2pq}x=[p−qa−ba+b]2pq
A) 1 B) 0 C) 12\frac{1}{2}21 D) (a−b)(a+b)\frac{(a-b)}{(a+b)}(a+b)(a−b) E) (a+b)(a−b)\frac{(a+b)}{(a-b)}(a−b)(a+b)
Si a2−bcbc+b2−acac+c2−abab=0\frac{a^2-bc}{bc} + \frac{b^2-ac}{ac} + \frac{c^2-ab}{ab} = 0bca2−bc+acb2−ac+abc2−ab=0; {a;b;c}⊂R\{a; b; c\} \subset \mathbb{R}{a;b;c}⊂R.
Halle
b+ca+c+ab+a+bc\frac{b+c}{a} + \frac{c+a}{b} + \frac{a+b}{c} ab+c+bc+a+ca+b
A) 2 B) 3 C) 6 D) 4 E) A ∨\lor∨ C
Sabiendo que: (a−b)(a−c)+(b−c)(b−a)+(c−a)(c−b)=2(a+b+c)2(a-b)(a-c) + (b-c)(b-a) + (c-a)(c-b) = 2(a+b+c)^2(a−b)(a−c)+(b−c)(b−a)+(c−a)(c−b)=2(a+b+c)2
Donde abc≠0abc \neq 0abc=0, halle el valor de:
a+3b+cac+b+3c+aab+c+3a+bbca−1+b−1+c−1 \frac{ \frac{a+3b+c}{ac} + \frac{b+3c+a}{ab} + \frac{c+3a+b}{bc} }{ a^{-1} + b^{-1} + c^{-1} } a−1+b−1+c−1aca+3b+c+abb+3c+a+bcc+3a+b
A) -1 B) -4 C) -9 D) -13 E) -15
Si ab+ba=1\frac{a}{b} + \frac{b}{a} = 1ba+ab=1 calcule el valor de:
(2(ab)+(ba)3)3+(2(ba)+(ab)3)3 \left( 2\left(\frac{a}{b}\right) + \left(\frac{b}{a}\right)^3 \right)^3 + \left( 2\left(\frac{b}{a}\right) + \left(\frac{a}{b}\right)^3 \right)^3 (2(ba)+(ab)3)3+(2(ab)+(ba)3)3
A) 0 B) 1 C) -1 D) 2 E) -2
Si y−1−x−1=(12)−1y^{-1} - x^{-1} = \left(\frac{1}{2}\right)^{-1}y−1−x−1=(21)−1 el valor reducido de
16x4y4+(x2−y2)2x4y2+x2y4\frac{16x^4y^4 + (x^2-y^2)^2}{x^4y^2 + x^2y^4} x4y2+x2y416x4y4+(x2−y2)2
será
A) 3 B) 2 C) 8 D) 4 E) 1
Sabiendo que x−1+y−1+z−1=0∧xyz≠0x^{-1} + y^{-1} + z^{-1} = 0 \land xyz \neq 0x−1+y−1+z−1=0∧xyz=0 señale el equivalente de:
x9+y9+z9−3xyz(x6+y6+z6)+6x3y3z3x6+y6+z6−3x2y2z23 \sqrt[3]{ \frac{ x^9 + y^9 + z^9 - 3xyz(x^6 + y^6 + z^6) + 6x^3y^3z^3 }{ x^6 + y^6 + z^6 - 3x^2y^2z^2 } } 3x6+y6+z6−3x2y2z2x9+y9+z9−3xyz(x6+y6+z6)+6x3y3z3
A) xyzxyzxyz B) x2+y2+z2x^2 + y^2 + z^2x2+y2+z2 C) x + y + z D) x3+y2+z3x^3 + y^2 + z^3x3+y2+z3 E) (x+y+z)3(x + y + z)^3(x+y+z)3
Si x+y+z=1x+y+z=1x+y+z=1 ; x3+y3+z3=4x^3+y^3+z^3=4x3+y3+z3=4. Calcule
1x+yz+1y+xz+1z+xy\frac{1}{x+yz} + \frac{1}{y+xz} + \frac{1}{z+xy} x+yz1+y+xz1+z+xy1
A) 3/2 B) 1 C) 3 D) -1 E) -2
Sabiendo que abc=1abc=1abc=1, el valor de:
aab+a+1+bbc+b+1+cac+c+1 \frac{a}{ab+a+1} + \frac{b}{bc+b+1} + \frac{c}{ac+c+1} ab+a+1a+bc+b+1b+ac+c+1c
A) 3 B) -1 C) 2 D) 1 E) 0
Si se cumple que 2x−1=2−x2x^{-1} = 2 - x2x−1=2−x.
Calcule el valor de [x9−(x4+x2+1)(x6+x3+1)]3\left[ x^9 - (x^4+x^2+1)(x^6+x^3+1) \right]^3[x9−(x4+x2+1)(x6+x3+1)]3
A) 0 B) 1 C) 2 D) x3x^3x3 E) 1x\frac{1}{x}x1
a+b+c=xa+b+c=x a+b+c=x
ab+bc+ac=4x2ab+bc+ac=4x^2 ab+bc+ac=4x2
abc=3x3abc=3x^3 abc=3x3
Determinar el valor de:
P=(a+b+ac)(a+2b+c)(2a+b+c)(a+b)(b+c)(a+c)P = \frac{(a+b+ac)(a+2b+c)(2a+b+c)}{(a+b)(b+c)(a+c)} P=(a+b)(b+c)(a+c)(a+b+ac)(a+2b+c)(2a+b+c)
A) 10 B) 5−x5-x5−x C) 3+x3+x3+x D) 7 E) 9
a+b+c=0a+b+c=0 a+b+c=0
ab+bc+ac=1ab+bc+ac=1 ab+bc+ac=1
abc≠0abc \neq 0 abc=0
Hallar el valor de:
(a3+b3+c3abc)2+a5+b5+c5abc\left(\frac{a^3+b^3+c^3}{abc}\right)^2 + \frac{a^5+b^5+c^5}{abc} (abca3+b3+c3)2+abca5+b5+c5
A) 4a4a4a B) -4 C) 4 D) 0 E) 4b4b4b
Si se sabe que:
(2m)8x+(2m)−8x=7∧0<m<12(2m)^{8x}+(2m)^{-8x}=7 \quad \land \quad 0<m<\frac{1}{2} (2m)8x+(2m)−8x=7∧0<m<21
P=(2m)−2x−(2m)2xP = (2m)^{-2x}-(2m)^{2x} P=(2m)−2x−(2m)2x
A) -2 B) 3 C) −12-\frac{1}{2}−21 D) 2 E) 1
(a+1)(a+b)=b;a≠0(a+1)(a+b)=b ; a \neq 0 (a+1)(a+b)=b;a=0
P=a3−3ab+b33P = \sqrt[3]{a^3-3ab+b^3} P=3a3−3ab+b3
A) 1 B) 2 C) -1 D) -2 E) 0
Siendo:
P=(A+B+C+D+E)(A+B−C−D−E)P = (A+B+C+D+E)(A+B-C-D-E) P=(A+B+C+D+E)(A+B−C−D−E)
Q=(A−B+C+D−E)(A−B−C−D+E)Q = (A-B+C+D-E)(A-B-C-D+E) Q=(A−B+C+D−E)(A−B−C−D+E)
Donde: P=Q∧P=Q \landP=Q∧ además: AB=43;C+D=323AB=\sqrt[3]{4} ; C+D=\sqrt[3]{32}AB=34;C+D=332 Halle: 'E'
A) 0,5 B) 4 C) 1 D) 23\sqrt[3]{2}32 E) 2
x=2abb2+1;y=2aba2+1x = \frac{2ab}{b^2+1} ; y = \frac{2ab}{a^2+1} x=b2+12ab;y=a2+12ab
Además:
(ab+1)(a+b)(ab−1)(a−b)=53;a;b>1\frac{(ab+1)(a+b)}{(ab-1)(a-b)} = \frac{5}{3} ; a;b>1 (ab−1)(a−b)(ab+1)(a+b)=35;a;b>1
P=a+x+a−xb+y+a−yP = \frac{\sqrt{a+x}+\sqrt{a-x}}{b+y+\sqrt{a-y}} P=b+y+a−ya+x+a−x
A) 2 B) 5 C) 8 D) 10 E) 1
ab+bc+ac=−8ab+bc+ac=-8 ab+bc+ac=−8
(a+b)2+(a+c)2+(b+c)2=16(a+b)^2+(a+c)^2+(b+c)^2=16 (a+b)2+(a+c)2+(b+c)2=16
P=a2b−1c−1+b2c−1a−1+c2a−1b−1P = a^2b^{-1}c^{-1}+b^2c^{-1}a^{-1}+c^2a^{-1}b^{-1} P=a2b−1c−1+b2c−1a−1+c2a−1b−1
A) -3 B) 3 C) 1 D) -2 E) 2
a=10+99+10−993a = \sqrt[3]{\sqrt{10+\sqrt{99}}+\sqrt{10-\sqrt{99}}} a=310+99+10−99
b=10+99−10−993b = \sqrt[3]{\sqrt{10+\sqrt{99}}-\sqrt{10-\sqrt{99}}} b=310+99−10−99
M=(a−b)(a5+b5)+(a+b)(a5−b5)M = (a-b)(a^5+b^5)+(a+b)(a^5-b^5) M=(a−b)(a5+b5)+(a+b)(a5−b5)
A) 15 B) 53 C) 125 D) 8 E) 0
A3+3A+6B2+2C23+A3+3A−6B2−2C23=B\sqrt[3]{A^3+3A+6B^2+2C^2}+\sqrt[3]{A^3+3A-6B^2-2C^2}=B 3A3+3A+6B2+2C2+3A3+3A−6B2−2C2=B
Calcular el valor de 'C' en:
C=A3+3A+6B2+2C23−A3+3A−6B2−2C23C = \sqrt[3]{A^3+3A+6B^2+2C^2}-\sqrt[3]{A^3+3A-6B^2-2C^2} C=3A3+3A+6B2+2C2−3A3+3A−6B2−2C2
A) 16 B) 32 C) 8 D) 2AB E) 4